To find the point that is symmetrical to the point (-1, -3) with respect to the line represented by the equation \(y - 2 = 12x + 8\), we first need to rewrite this equation in slope-intercept form (y = mx + b).
Starting with the given equation:
\[ y - 2 = 12x + 8 \] We add 2 to both sides to obtain:
\[ y = 12x + 10 \]
This is the line with a slope of 12 and a y-intercept at (0, 10).
Next, we want to find the perpendicular distance from the point (-1, -3) to the line. The slope of the line is 12, so the slope of the line perpendicular to it is the negative reciprocal, which is \(-\frac{1}{12}\).
Now we can find the equation of the line that passes through the point (-1, -3) with this perpendicular slope. Using the point-slope form of the equation of a line \(y - y_1 = m(x - x_1)\):
\[ y + 3 = -\frac{1}{12}(x + 1) \]
Expanding this, we have:
\[ y + 3 = -\frac{1}{12}x - \frac{1}{12} \] Subtracting 3 from both sides gives:
\[ y = -\frac{1}{12}x - \frac{1}{12} - 3 \]
To combine the constant terms:
\(-3 = -\frac{36}{12}\), so:
\[ y = -\frac{1}{12}x - \frac{1 + 36}{12} = -\frac{1}{12}x - \frac{37}{12} \]
Now we have the equation of the perpendicular line, which we can use to find the intersection point with the original line \(y = 12x + 10\). We will set the two equations equal to each other:
\[ 12x + 10 = -\frac{1}{12}x - \frac{37}{12} \]
To eliminate the fraction, we multiply the entire equation by 12:
\[ 12(12x + 10) = -x - 37 \] That simplifies to:
\[ 144x + 120 = -x - 37 \]
Adding \(x\) to both sides gives:
\[ 145x + 120 = -37 \]
Subtracting 120 from both sides:
\[ 145x = -37 - 120 \]
This results in:
\[ 145x = -157 \] Dividing both sides by 145 gives:
\[ x = -\frac{157}{145} \]
Now we can substitute this value back into the equation of the original line to find \(y\):
\[ y = 12(-\frac{157}{145}) + 10 \] Calculating this gives:
\[ y = -\frac{1884}{145} + \frac{1450}{145} = \frac{-1884 + 1450}{145} = \frac{-434}{145} \]
The intersection point is \((- \frac{157}{145}, - \frac{434}{145})\).
Now, we find the point symmetrical to (-1, -3). The midpoint between (-1, -3) and the symmetrical point \((x', y')\) is the intersection point we just found:
\[ \left(\frac{-1 + x'}{2}, \frac{-3 + y'}{2}\right) = \left(-\frac{157}{145}, -\frac{434}{145}\right) \]
Thus, we can set up the equations:
- \(\frac{-1 + x'}{2} = -\frac{157}{145}\)
- \(\frac{-3 + y'}{2} = -\frac{434}{145}\)
Solving the first equation:
\[ -1 + x' = -\frac{314}{145} \] \[ x' = -\frac{314}{145} + 1 = -\frac{314}{145} + \frac{145}{145} = -\frac{169}{145} \]
Now for \(y'\):
\[ -3 + y' = -\frac{868}{145} \] \[ y' = -\frac{868}{145} + 3 = -\frac{868}{145} + \frac{435}{145} = -\frac{433}{145} \]
Therefore, the point that is symmetrical to (-1, -3) with respect to the line \(y - 2 = 12x + 8\) is:
\[ \left(-\frac{169}{145}, -\frac{433}{145}\right) \]